Compacted soil 18 kN/m³, saturation

Given

$\gamma_{d} = 18$ kN/m³, $w = 15\%$, $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

Degree of saturation

$$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{18} - 1 = 0.444$$ $$S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.444} = 0.895 = 89.5\%$$

Further compaction

The maximum value the dry unit weight can reach at the same water content is when all air is expelled ($S = 100\%$), on the ZAV line:

$$\gamma_{d,max} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.15 \times 2.65} = \frac{25.99}{1.3975} = 18.60\ \text{kN/m}^3$$

Answer: $S = 89.5\%$; the soil can at most be compacted to 18.60 kN/m³ (in practice slightly less, since some air always remains).